Calculate the energy that can be obtained from 1 kg of water through the fusion reaction ^{2}H + ^{2}H → ^{3}H + p. Assume that 1.5 × 10^{−2}% of natural water is heavy water D_{2}O (by number of molecules) and all the deuterium is used for fusion.

(Use Mass of proton m_{p} = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_{n} = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c^{2},1 u = 931 MeV/c^{2}.)

#### Solution

18 g of water contains `6.023 xx 10^23` molecules.

`∴ "1000 g of water" = (6.023 xx 10^23 xx 1000)/18 = 3.346 xx 10^25` molecules

`% "of deuterium" = 3.346 xx 10^25 xx 0.015/100 = 0.05019 xx 10^23`

Energy of deuterium = `30.4486 xx 10^25`

`= [2 xx m(""^2H) - m(""^3H) - m_p]c^2`

`= (2 xx 2.014102 "u" - 3.016049 "u" - 1.007276 "u")c^2`

`= 0.004879 xx 931 "MeV"`

`= 4.542349" MeV"`

`= 7.262 xx 10^-13 "J"`

Total energy = `0.05019 xx 10^23 xx 7.262 xx 10^-13 "J"`

= 3644 MJ